# I don't know how to find the normal force on this problem

1. A m = 3.0 kg wood box slides down a vertical wood wall while you push on it at a θ = 49° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

2. Fk-Mk*n

3. i know there are 4 forces acting on the object. gravity, normal force from the wall, kinetic friction, and the force that i applied. but i do not know how to set up the equation to solve this problem

Do you know how to draw a free body diagram?

yea i did one but the n is on the x axis so that means i need to know the F to get the normal force right?

First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using newton's second law for net force in the x and y directions?

First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using newton's second law for net force in the x and y directions?

o i am sorry this question some how expected us to find the coefficient of kinetic friction on wood which is 0.2. and yes

tiny-tim
Homework Helper
Welcome to PF!

Hi xstetsonx! Welcome to PF! i need to know the F to get the normal force right?

No, you don't need to know F, you can just call it F.

For constant speed, you need zero acceleration, which means zero net force in both the x and y direction.

So call the force F, find N, and then find µN. so the total force will be
mK*Fcos49-mg+Fsin49?

tiny-tim
Homework Helper
so the total force will be
mK*Fcos49-mg+Fsin49?

Yes, except I'm not sure about that m at the beginning.

You should get

$$F_{app} - F_{k} - mg\sin(\theta) = 0$$
$$n - mg\cos(\theta) = 0$$

The algebra's up to you.

Edit:

so the total force will be
mK*Fcos48-mg+Fsin48?

Yes, that's correct, assuming F is the force of gravity.